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If $n>2$ and $k$ is positive integer, then there is no positive integer $m$ satisfy that $$k(k+1)\cdots (k+n-1)=m^n\, ?$$

I tried to prove this problem, but I don't know how to prove it. I know that every product of consecutive integers is not a power, but I want to simple proof of my question. Thanks for any help.

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  • $\begingroup$ @LoganMaingi I do not watch any anime. However, this problem was inspired by Nisekoi. I saw this problem from some facebook group (this post is writed in Korean!), and this post say that this problem is appear in the anime Nisekoi. $\endgroup$ – Hanul Jeon Feb 9 '14 at 8:26
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This solution is false. (I cannot delete an accepted answer.)


Hint: Because you have the same number of terms on both sides, $m$ must be very close to $k + \frac{n}{2} $.

The $n$ odd case is pretty easy, you should be quickly able to show that

$$(k + \frac{n-1}{2} -1 )^n < k(k+1) \ldots (k+n-1) < ( k + \frac{n-1}{2} ) ^n, $$

hence there is no integer $m$ that will work. The RHS works by pairing up terms of the form $(k+ \frac{n-1}{2} - i)(k+ \frac{n-1}{2} + i ) < (k + \frac{n-1}{2} ) ^2$, and the LHS works by pairing up terms of the form $(k+ \frac{n-1}{2} - i)(k+ \frac{n-1}{2} + i +1 ) > (k + \frac{n-1}{2} -1 )^2 $

Do the same for $n$ even, with a slightly different inequality.

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  • $\begingroup$ Why is this true: $(k+ \frac{n-1}{2} - i)(k+ \frac{n-1}{2} + i +1 ) > (k + \frac{n-1}{2} -1 )^2$? If $i=k+\frac{n-1}2-1$, then this is not always true, or am I missing something? $\endgroup$ – awllower Jun 26 '18 at 11:06
  • $\begingroup$ @awllower Note that $i$ is at most $ \frac{n-1}{2}$, so your case doesn't work. However, with $i = \frac{n-1}{2}$, it is often not true that $k (k+n) > (k + \frac{n-1}{2} -1)^2 $. $\endgroup$ – Calvin Lin Jul 27 '18 at 15:53
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I think I found a simple solution, will need someone to verify.

suppose $m$ exists such that $k(k+1)⋯(k+n−1)=m^n$

then $k < m < k + n - 1$

$gcd(m,m+1) = 1$

$m+1$ divides the LHS but doesn't divide the RHS

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This, and variations of it appears in:

P. Erdős and J. L. Selfridge, Product of Consecutive integers is never a power, Illinois J. of Math. 19(1975) 292- 301.

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  • $\begingroup$ That is a much stronger result (with high power machinery) than what is needed. $\endgroup$ – Calvin Lin Feb 6 '14 at 17:23
  • $\begingroup$ This is what I thought of immediately, but it would be better as a comment (unless accompanied by a synopsis of some of the proof). $\endgroup$ – robjohn Dec 19 '14 at 8:21

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